Originally posted by teknofreaker:
Thanks!
http://mathforum.org/dr.math/faq/faq.interest.html#install
Example: Suppose you finance your car with a loan of $12000 at an interest rate of 11% for four years, and make equal payments monthly. How much will your payments have to be? Here the parameters are principal P = $12000, interest rate i = 0.11, number of years n = 4, and number of periods per year q = 12. Then the monthly car payment M is given by
M = Pi/[q(1-[1+(i/q)]-nq)],
= ($12000)(0.11)/[(12)(1-[1+(0.11/12)]-(4)(12))],
= $110/(1-1.009166666...-4
,
= $310.15.
How much will be owed at the end of two years? Here we have a monthly payment of M = $310.15, and the number of periods k = 2q = 24, and compute the amount due A at that time.
A = (P-Mq/i)(1+[i/q])k + Mq/i,
= ($12000-[$310.15][12]/0.11)(1+[0.11/12])24 +
($310.15)(12)/0.11,
= -$27180.264935 + $33834.545454,
= $6654.28.
How much interest will have been paid in total?
I = Mnq - P,
= ($310.15)(4)(12) - $12000,
= $14887.20 - $12000,
= $2887.20.
